A note on products of quadratic matrices of singular type

TL;DR

This paper offers a new proof of a theorem related to the products of idempotent and square-zero matrices, and characterizes products of singular quadratic matrices using existing results.

Contribution

It presents an alternative, self-contained proof of Botha's theorem and provides a comprehensive characterization of products of singular quadratic matrices.

Findings

Proof of Botha's theorem is simplified and clarified.

Characterization of products of singular quadratic matrices established.

Enhanced understanding of matrix product structures in linear algebra.

Abstract

I provide an alternate and self-contained proof of Botha's theorem on products of idempotent and square-zero matrices where the product contains two square-zero factors, and provide a conclusive characterisation of products of singular quadratic matrices based on previous results.

Equations

$$J_k(0)=\left[\begin{array}{cc}0& 0\\ I_{k-1}& 0\end{array}\right].$$

$$(\text{R}(G)\cap \text{N}(G))+\text{N}(F)\subseteq \text{N}(G),$$

$$\dim (\text{R}(G)\cap \text{N}(G))+\text{n}(F)-\dim (\text{R}(G)\cap \text{N}(G)\cap \text{N}(F))\le \text{n}(G).$$

$$\text{R}(G)\cap \text{N}(F)=\text{R}(G)\cap \text{R}(F)\cap \text{N}(F)\subseteq \text{R}(F)\cap \text{N}(F),$$

$$\dim (\text{R}(F)\cap \text{N}(F))\le \text{n}(F)-\text{r}(F).$$

$$n(F)-{\text{n}}_0(G)\le \text{n}(F)-\text{r}(F),$$

$$\text{r}(G)\le \text{r}(F)\le {\text{n}}_0(G).$$

$$\text{R}(G)\cap \text{N}(F)$$

$$\text{r}(G)\le {\text{n}}_0(G)+\dim (G(\mathcal{W})),$$

$$\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right].$$

$$F_1=\left[\begin{array}{c}{I}_{k-2}\\ 0\end{array}\right],$$

$$\left[\begin{array}{cc}0& 0\\ I_{k-1}& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ e_{k-1}& I_{k-1}\end{array}\right]\left[\begin{array}{cc}{e}_{k-1}^T& 0\\ F_1& 0\end{array}\right].$$

Full text

1 Introduction

A singular quadratic matrix which is not the zero matrix has minimum polynomial $x^{2}-cx$ (where $c$ is some scalar over a field), and is therefore either square-zero, or a scalar multiple of an idempotent matrix. In this article I aim to summarise previous results on products of such matrices in one general statement, presented here as corollary 7.

I want to start however, by presenting an alternate proof to one of the results first proved by Botha [4]. In this article I aim to present additional insight into that investigation through a self-contained alternate proof which might provide further insight and aid further research into products of quadratic matrices.

I will now fix some notation. The set of order $n\times n$ square matrices over a field $\mathscr{F}$ is indicated as $M_{n}(\mathscr{F})$. Matrices are generally indicated by capitals, and vectors in lower case. The null space of a matrix $G$ is indicated as $\operatorname{N}(G)$ and its range as $\operatorname{R}(G)$, the corresponding dimensions of these subspaces are indicated as $\operatorname{n}(G)$ (nullity of $G$) and $\operatorname{r}(G)$ (rank of $G$) respectively.

Definition. $\text{n}_{0}(G)=\text{n}(G)-\dim(\text{R}(G)\cap\text{N}(G))$.

I denote the simple Jordan block with characteristic value zero and order $k\times k$ as

[TABLE]

It is easy to see that $\operatorname{n}_{0}(G)$ is the number of blocks $J_{1}(0)$ in the rational canonical form of $G$.

A vector space is generally indicated in capitals by calligraphic font. The vector space of $n$-dimensional (column) vectors are indicated as $\mathscr{F}^{n}$, and the standard basisvector with a 1 in entry $i$ and zeros elsewhere is indicated as $e_{i}$.

Well known previous results employed in the proof include

Grassmann’s theorem: Let $\mathcal{V}$ be a vector space over a field $\mathscr{F}$ and let $\mathcal{W}$ and $\mathcal{Y}$ be subspaces of $\mathcal{V}$ satisfying the condition that $\mathcal{W}+\mathcal{Y}$ has finite dimension. Then $\dim(\mathcal{W}+\mathcal{Y})=\dim(\mathcal{W})+\dim(\mathcal{Y})-\dim(\mathcal{W}\cap\mathcal{Y})$. 2. 2.

Sylvester’s theorem: Let $A,B$ be matrices over $\mathscr{F}$ such that $A$ has $n$ columns and $B$ has $n$ rows. Then (i) $\operatorname{n}(AB)\leq\operatorname{n}(A)+\operatorname{n}(B)$; $\operatorname{r}(A)+\operatorname{r}(B)-n\leq\operatorname{r}(AB)\leq\min\{\operatorname{r}(A),\operatorname{r}(B)\}$. 3. 3.

Fitting’s lemma: $G\in M_{n}(\mathscr{F})$ is similar to $N\oplus B$ where $N$ is nilpotent and $B$ is non-singular.

Recent previous results that are employed:

$H\in M_{n}(\mathscr{F})$ is a product of $k\geq 1$ idempotent matrices (over $\mathscr{F}$) of nullities $n_{1},\ldots,n_{k}$ respectively if and only if $\operatorname{n}(H)\geq n_{i}\geq 0$ and $n_{1}+\cdots+n_{k}\geq\operatorname{r}(I-H)$. [1] [2] 2. 2.

$F\in M_{n}(\mathscr{F})$ is a product of two square-zero matrices $Z_{1},Z_{2}$ if and only if $\operatorname{r}(F)\leq\operatorname{n}_{0}(F)$, and whenever $F$ is such a product the ranks of $Z_{1},Z_{2}$ can be arbitrarily chosen subject to $\operatorname{r}(G)\leq\operatorname{r}(Z_{1}),\operatorname{r}(Z_{2})\leq n/2$. [3, theorem 3]

Finally note that, up to similarity, the ordering of factors in a product consisting entirely of idempotent and square-zero matrices with prescribed nullities may be chosen arbitrarily, due to the fact that the idempotent and square-zero properties are preserved by similarity, and furthermore a product $AB$ is similar to $B^{T}A^{T}$ (its transpose) for any two square matrices $A,B$ of the same order. We may therefore generalize the results below to include the case where $G$ consists of any ordering of the factors presented.

2 Products with two square-zero factors

Theorem 1**.**

Let $G\in M_{n}(\mathscr{F})$ and $n_{1},\ldots,n_{k},n_{Z_{1}},n_{Z_{2}}\in\mathbb{N}$. The following two statements are equivalent.

$G=E_{1}\ldots E_{k}Z_{1}Z_{2}$ where $E_{i}^{2}=E_{i}$ and $\operatorname{n}(E_{i})=n_{i}$ for each $i\in\{1,\ldots,k\}$, and $Z_{1}^{2}=Z_{2}^{2}=\mathbf{0}$ and $\operatorname{n}(Z_{1})=n_{Z_{1}}$ and $\operatorname{n}(Z_{2})=n_{Z_{2}}$. 2. 2.

1. (i)

$\operatorname{r}(G)\leq n_{1}+\cdots+n_{k}+\operatorname{n}_{0}(G)$, 2. (ii)

$n_{1},\ldots,n_{k},n_{Z_{1}},n_{Z_{2}}\leq\operatorname{n}(G)$, 3. (iii)

$n_{Z_{1}},n_{Z_{2}}\geq n/2$.

Lemma 2**.**

Let $G=HF$ where $H$ and $F$ are arbitrary square matrices. Then $\dim(\text{R}(G)\cap\text{N}(F))\geq\text{n}(F)-\text{n}_{0}(G)$.

Proof.

[TABLE]

and so by Grassmann’s theorem

[TABLE]

Now since $N(F)\subseteq N(G)$, the desired result follows directly from (1). $\square$

Lemma 3**.**

Let $G=HF$ where $H$ is an arbitrary square matrix and $F=Z_{1}Z_{2}$, where $Z_{1}^{2}=Z_{2}^{2}=\mathbf{0}$. If $\text{R}(G)\cap\text{N}(F)=\text{R}(G)\cap\text{R}(F)\cap\text{N}(F)$ then $\text{r}(G)\leq\text{n}_{0}(G)$.

Proof. We have

[TABLE]

and since $F$ is the product of two square-zero matrices we have by [3, theorem 3]

[TABLE]

It follows by lemma 2 that

[TABLE]

and therefore

[TABLE]

$\square$

Remark: Notice that the result above includes the case where $\text{R}(G)\cap\text{N}(F)=\{0\}$.

Proposition 4**.**

Let $G=HF$ where $H$ is an arbitrary square matrix and $F=Z_{1}Z_{2}$, where $Z_{1}^{2}=Z_{2}^{2}=\mathbf{0}$. Then $\operatorname{r}(G)\leq\operatorname{n}_{0}(G)+\operatorname{r}(I-H)$.

Proof. If $\text{R}(G)\cap\text{N}(F)=\text{R}(G)\cap\text{R}(F)\cap\text{N}(F)$ then the result follows directly from lemma 3, so suppose that $\text{R}(G)\cap\text{R}(F)\cap\text{N}(F)\subset\text{R}(G)\cap\text{N}(F)$. Then there exists a subspace $\mathcal{W}$ such that $\dim(G(\mathcal{W}))=\dim(\mathcal{W})>0$ and

[TABLE]

Now by a similar argument as employed in the proof of lemma 3 it is easy to show

[TABLE]

and it remains to show that $\dim(G(\mathcal{W}))\leq\text{r}(I-H)$.

Let $\{w_{1},w_{2},\ldots,w_{k}\}$ be a basis for $\mathcal{W}$, then since $\mathcal{W}\cap\text{N}(G)=\{0\}$ it follows that $\{Gw_{1},Gw_{2},\ldots,Gw_{k}\}$ is a linearly independent set, and since $\text{R}(F)\cap G(\mathcal{W})=\{0\}$ we then also have $\{(F-G)w_{1},(F-G)w_{2},\ldots,(F-G)w_{k}\}$ is linearly independent. Now $F-G=(I-H)F$ and it follows that $\{(F-G)w_{1},(F-G)w_{2},\ldots,(F-G)w_{k}\}\subseteq\text{R}(I-H)$, which yields the desired result. $\square$

Lemma 5**.**

Let $k\geq 2$, then $J_{k}(0)=EF$ where $\text{N}(F)=\text{N}(J_{k}(0))$ and $\text{n}_{0}(F)=1$ and $E^{2}=E$ and $\text{n}(E)=1$.

Proof. If $k=2$:

[TABLE]

If $k>2$, let

[TABLE]

then:

[TABLE]

$\square$

Remark: Notice that $n_{0}(J_{k}(0))=0$ for any integer $k\geq 2$.

Corollary 6**.**

Let $J=J_{k_{1}}(0)\oplus\cdots\oplus J_{k_{m}}(0)$ where $k_{i}\geq 2$. Then $J=EF$ where $E^{2}=E$ and $\operatorname{N}(F)=\operatorname{N}(J)$ and $0\leq\text{n}_{0}(F)\leq\text{n}(E)\leq m$.

Proof. Let $0\leq s\leq m$, and let $J_{k_{i}}(0)=E_{i}F_{i}$ as in the lemma above for $1\leq i\leq s$. Furthermore, choose $E_{i}=I_{k_{i}}$ or $E_{i}=0\oplus I_{k_{i}-1}$ for $s<i\leq m$, so that $J_{k_{i}}(0)=E_{i}J_{k_{i}}(0)$ in this case.

Then $J=EF$ with $F=F_{1}\oplus\cdots\oplus F_{s}\oplus J_{k_{s+1}}(0)\oplus\cdots\oplus J_{k_{m}}(0)$ and $E=E_{1}\oplus\cdots\oplus E_{m}$. $\square$

Proof of theorem 1. Suppose point number 1 in the statement of the theorem is true. Then (ii) and (iii) follow easily, and (i) follows from proposition 4 and [1].

Suppose point number 2 in the statement of the theorem holds. By the final paragraph of the introduction we may assume without loss of generality that $n_{1}\leq\cdots\leq n_{k}$.

By Fitting’s lemma $G$ is similar to $N\oplus B$ where $N$ is nilpotent and $B$ is invertible. Without loss of generality we can assume that $N=\mathbf{0}_{\text{n}_{0}(G)}\oplus J$ where $J=J_{k_{1}}(0)\oplus\cdots\oplus J_{k_{m}}(0)$ with $k_{i}\geq 2$ and $m=\dim(\text{R}(G)\cap\text{N}(G))$.

Now $\mathbf{0}_{\text{n}_{0}(G)}=H_{1}\mathbf{0}_{\text{n}_{0}(G)}$ where $H_{1}$ is any idempotent matrix with $0\leq\operatorname{n}(H_{1})\leq\text{n}_{0}(G)$.

Let $s=\max\{\text{r}(G)-\text{n}_{0}(G),0\}$. It then follows that $s\leq n_{1}+\cdots+n_{k}$ and the corollary above shows that $J=H_{2}F_{2}$ where $H_{2}$ is idempotent, $\operatorname{N}(F_{2})=\operatorname{N}(J)$ and $\text{n}_{0}(F_{2})=s\leq\operatorname{n}(H_{2})\leq m$.

Finally we have $B=I_{\text{r}(G)}B$.

Now $E=H_{1}\oplus H_{2}\oplus I_{\text{r}(G)}$ is idempotent. Notice that by the choice of $H_{1}$ and $H_{2}$, we can specify $\operatorname{n}(E)$ arbitrarily subject to $s\leq\operatorname{n}(E)\leq\operatorname{n}(G)$ and therefore let $\operatorname{n}(E)=\max\{n_{k},s\}$. Now since $\text{n}(E)=\text{r}(I-E)$, the result [1] shows that $E$ can be written as $E_{1}\ldots E_{k}$.

Furthermore $F=\mathbf{0}_{\text{n}_{0}(G)}\oplus F_{2}\oplus B$ is such that $\text{n}_{0}(F)=\operatorname{n}_{0}(G)+s\geq\text{r}(G)=\text{r}(F)$. By theorem 3 of [3] it follows that $F=Z_{1}Z_{2}$. $\square$

3 Products of quadratic matrices of singular type

Corollary 7**.**

Let $G\in M_{n}(\mathscr{F})$ and $n_{1},\ldots,n_{k},m_{1},\ldots,m_{l}\in\mathbb{N}$ and $c_{1},\ldots,c_{k}$ be nonzero scalars in $\mathscr{F}$. Set $c=(c_{1}\ldots c_{k})^{-1}$. The following two statements are equivalent.

$G=(c_{1}E_{1})\ldots(c_{k}E_{k})Z_{1}\ldots Z_{l}$ where $E_{i}^{2}=E_{i}$ and $\operatorname{n}(E_{i})=n_{i}$ for each $i\in\{1,\ldots,k\}$, and $Z_{j}^{2}=\mathbf{0}$ and $\operatorname{n}(Z_{j})=m_{j}$ for each $j\in\{1,\ldots,l\}$. 2. 2.

1. (i)

$n_{1},\ldots,n_{k},m_{1},\ldots,m_{l}\leq\operatorname{n}(G)$, 2. (ii)

$m_{1},\ldots,m_{l}\geq n/2$, 3. (iii)

furthermore if $l=0$ then $\operatorname{r}(I-cG)\leq n_{1}+\cdots+n_{k}$,

if $l=1$ then $\dim(\operatorname{R}(G)+\operatorname{N}(G))\leq n_{1}+\cdots+n_{k}+m_{1}$,

if $l=2$ then $\operatorname{r}(G)\leq n_{1}+\cdots+n_{k}+\operatorname{n}_{0}(G)$.

Proof. Note that $G=(c_{1}E_{1})\ldots(c_{k}E_{k})Z_{1}\ldots Z_{l}$ is equivalent to $cG=E_{1}\ldots E_{k}Z_{1}\ldots Z_{l}$. Now the corollary follows easily by combining the results presented in [1], [2], [4], and the preceding sections, with the fact that for any subspace $S$ we have $cS=\{cv:v\in S\}=S$. $\square$